## Sandwich Panel Flexure Stress Equation Derivation

Look at the equation in the picture above, did you ever wonder how this **sandwich panel flexure** stress equation derivation came about?

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I will try to derive this equation in this post. Ready? Let’s begin…

Sandwich Panel Flexure Stress Equation Derivation:

The flexure stress equation is a derivative of the basic plate bending stress equation as shown in the figure above.

For a plate, it is simply [math] Mc/I [/math]:

Where:

- ‘M’ is the applied moment (in-lb)
- ‘c’ is the extreme fiber distance from the neutral axis, where the bending stresses are the highest (in)
- ‘I’ is the effective area moment of inertia of the cross section ([math]in^4[/math])

In case of sandwich panels, we start with the same equation. However, the main underlying assumptions are as follows:

- Only the facings or the skins resist the lion’s share of the induced bending stresses
- The core mainly keeps the skins apart so they cannot move relative to each other
- The effective area of the skin is used in the area moment of inertia calculations along with the standard parallel axis theorem
- Very small quantities in the area moment of inertia calculation are zeroed out
- Skin thickness [math]t_f[/math] << Core [math]t_c[/math], usually in the order of 1/100 ths of an inch

**OK so let’s start with the basics:**

- The area of the facings or skins is [math] (t_f) * (b) in^2 [/math]
- The effective thickness of the sandwich panel, which is the mid plane distance between the facings, is [math] h = (t_f + t_c) [/math] [math] in [/math]

Basic Bending Stress Equation:

**[math]S = Mc/I[/math] —> (1)**

But now:

**[math]c = h/2[/math] —> (2)**

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Normally a rectangular I beam flange area moment of inertia [math] I = b*h^3/12+A*d^2 [/math]

But now:

b = panel width (in)

[math]h = t_f[/math] (in)

d = h/2 (in), the mid plane distance of the skin from the neutral plane

Therefore for teh two skins combined:

[math]A = b*t_f[/math] (each skin’s c/s area)

[math] I = 2*b*(t_f)^3/12 + 2*b*t_f*(h/2)^2 [/math]

[math] I = b*(t_f)^3/6 + b*t_f*(h)^2/2 [/math]

However, the first part of the RHS side of the above equation becomes a very tiny number. **The significance of this is that the skins themselves are basically incapable of resisting much bending as standalone parts of the layup, the crucial part is the separation that the core provides.** Read more about it **here**.

For example:

If [math]t_f = 0.02″[/math], then

[math] t_f^3/6 = 2.67E-8[/math] => 0 (approaches zero)

So the first part is zeroed out, hence:

**Effective [math] I = b*t_f*h^2/2 [/math] —> (3)**

Substituting (2) and (3) into the basic bending stress equation (1):

[math] S_f = Mc/I = M*(h/2)/(t_f*b*h^2/2)[/math]

### [math] S_f = M/(h*t_f*b)[/math]

**Alternative Approach:**

Another simpler approach is to use the force couple approach.

All the known quantities such as M, h etc. are the same. But this time we will use a section cut at the middle of the panel.

In this method, we cut a section through the middle of the long beam test coupon panel.

A force couple or resisting torque is developed within the skins, acting in the plane of the skins. Thus induces compressive stresses in the top skin and tensile stresses in the bottom skin.

The force is resisted by the cross sectional area of the skins (tf*b), we can now use the standard S = P/A formula to calculate the facing stresses.

Therefore:

### [math] S_f = (M/h)/(t_f*b)) [/math]

### ==> [math] S_f = M/(h*t_f*b)[/math]

So there you have it, hope you learnt something new and interesting today. Make sure you comment below the post and let me know what you think…

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